limits (2 Viewers)

[danif]

hey i posted this in the asymptotes thread but no one replied... =P so i figured i might start a thread - so appologies if you've read this:


i have a hypothetical (dontcha just love those)

when finding oblique asymptotes - you divide by the highest power of x in the denominator correct? what if the highest power is in the numerator? what do you do then? i.e. so you dont get 1/x = 0 as lim--> 00 (that's infinity) e.g:

x^3+ 3x^2 + 2x - 5
-------------------------
4x^2 - 2x^2 + 9

N.B: that is not an actual example so if the answer isnt pretty then you know why =)


thanks =)

 

[Managore]

diving all by x^2 we get (presuming the 2nd x^2 on the bottom is spose to be an x^1)

x + 3
-------
4

and that would be the answer, no?

 

[danif]

aaah, so then you sub in x= 00 ?

p.s. like i said this is not an actual example i.e. i just made it up as a hypothetical

 

[Managore]

umm divide by x^2, then sub in x=infinity, which reduces everything to basically 0, for all units which have a greater power of x on the denominator. 4x^2/x^2 of course cancels to 4, and 3x^2/x^2 cancels to 3. I left the x there as when x goes to infinity, it stays as x.. if that makes sense. It's just how I do it

 

[BillyMak]

If the highest power is in the numerator you can use either partial fractions (if you do 4-unit) or long division to simplify it down to an expression where the highest power is not on the numerator.

 

[Managore]

For that example, what would you get?

 

[BillyMak]

I'm not going to bother trying

We have the 3-unit test tomorrow and I am only attempting things that could be in the exam. The chances of a chestion like that being in the exam are very slight.